Monty Hall Problem
The Monty Hall problem has quite an unintuitive solution. For those who have not seen the problem before, you can see a reel describing it below:
Here is the problem in text form:
You’re on a game show, and you’re given the choice of three doors:
Behind one door is a car; behind the others, goats. Let us assume you want a car!
You pick a door, say number 1, and the host, who knows what’s behind the doors, opens another door, say number 3, which has a goat - he makes sure to open the door which has a goat behind it.
He then says to you, “Do you want to pick door number 2?” Is it to your advantage to switch your choice from door number 1 to door number 2?
At first glance, the solution to this problem seems like it shouldn’t matter whether you switch or not. It must be equally likely that the car is behind door 1 as behind door 2. However, that is not the case, however unintuitive it may seem. I claim that you should switch - there is a 2/3 chance that the goat is behind door 2 and only a 1/3 chance it is behind door 1. Before thinking about this as a probability problem, convince yourself through an empirical approach:
Approaching this problem empirically
Find a partner to play this game with you - decide which of you is the host and which is the player. Take three cards, two jacks representing the goats and a queen representing the car. First, the host should shuffle and place the cards on a table. Next, the player should pick a card - do not open it. Now, the host can look at the cards the player did not pick and choose a Jack and open it. The player can now choose to stay or to change. Keep track of what happened systematically. If you do this enough times, you will see that the player wins 2/3 of the times they switch and 1/3 of the time they do not. Run the game at least 100 times.
Is the host conveying information?
While doing the above experiment might be helpful in convincing yourself of the solution I suggested above, it probably does very little to help you understand why this is happening.
As Daniel Kahneman lays out in his research and book, Thinking Fast and Slow, probability is notoriously hard for humans to understand.
One way to develop an intuition for this problem is to think about how information changes with each step (I am using the term ‘information’ here in the colloquial sense). Specifically, focus on the following sentence:
You pick a door, say number 1, and the host, who knows what’s behind the doors, opens another door, say number 3, which has a goat - he makes sure to open the door which has a goat behind it.
The host has some information you do not have - he knows where the goats and cars are. In what he is doing, is he conveying any of that information to you? If he is, that may change the probabilities. To see whether he is, let us think about a related problem:
Increasing the number of doors
Many problems are easier to see a solution to by decreasing the number of objects we are dealing with. However, in this case, you will see that the opposite is the case. Let us take a situation with 100 doors, with one car, and 99 goats.
Suppose you pick one door, say door 48. What is the probability of you getting the car? Well, 1/100. Suppose the host, who knows where the car is, opens 98 other doors, all of which have goats behind them. You are left with door 48, which you originally picked, and, say, door 65, which the host left closed after opening 98 other doors.
Given that the host filtered the 99 remaining doors to 1, and you had initially selected 1 door out of 100 without any information, would you switch to door 65 or stay on door 48?
I’m guessing that you would agree in this case, it makes sense to switch - the host is conveying information to you by filtering the remaining doors. You did not use any information when selecting your original choice.
So, how does this translate into three doors?
Returning to three doors
In the case of three doors, the host is doing something similar - he is filtering the remaining two doors down to one.
How do we think about what is going on? Here are two ways
Focusing on the original choice
You would probably agree that the probability of your original choice being the car is 1/3. That means the probability that the car is in one of the other two doors is 2/3. Notice that nothing has changed about your original choice in opening one of the other doors. So, the probability of a car in your initial choice of door remains 1/3.
Where does the remaining 2/3 probability go? Well, the probability of the car being in the door opened by the host has become zero. So, it must be the case that the probability of the car being in the unopened door you did not select will be 2/3. This is because we know that there is a car behind one of the doors. So, the total probability across all the doors must be 1 or 100%.
Listing out all the possibilities
If the above argument doesn’t work for you, let us write out the possibilities in detail. Without loss of generality, we can say that you pick door 1 every time.
Behind door 1, there is either a car or a goat. If there is a car, there will be a goat behind doors 2 and 3. If there is a goat, the car is behind door 2 or 3.
In the first case, you are wrong to switch. However, you are right to switch in the second and third cases since the host will open door 3 in the second case and door 2 in the third case. So, if you switch every time, you will be right 2/3 of the time and wrong 1/3 of the time, which is the claim I made in the beginning.
To Conclude
The crux of this problem is information flow between the host, who knows the positions of the cars and goats, and you, the player. The host provides at least some of that information by filtering the remaining doors.
When thinking that the probability is 50-50, you are acting like you had a choice between two doors, one with a goat behind it and one with a car behind it. There was never a third door that the host filtered.
When I saw this problem for the first time, I had a strong intuition that the person who told me that it is better to switch was wrong. I was so convinced that I took a bet that, over 100 tries, the distribution which door the car was behind would be around 50-50. After that evening, I spent a lot of time convincing myself that switching makes sense. Do the above arguments persuade you? Do you have alternative, intuitive arguments for why the probabilities are this way? Please share below.